# Calculate L^(-1){(1)/(s^2(s^2+a^2))} using the convolution theorem L{f ** g}=L{f} * L{g}

Calculate ${\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}\left({s}^{2}+{a}^{2}\right)}\right\}$ using the convolution theorem $\mathcal{L}\left\{f\ast g\right\}=\mathcal{L}\left\{f\right\}\cdot \mathcal{L}\left\{g\right\}$
My approach:
$f\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}}\right\}=t$
$g\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}+{a}^{2}}\right\}=\frac{1}{a}\mathrm{sin}\left(at\right)$
${\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}\left({s}^{2}+{a}^{2}\right)}\right\}=f\ast g={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(t-\tau \right)g\left(\tau \right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau =\frac{1}{a}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left(t-\tau \right)\mathrm{sin}\left(a\tau \right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau$
but the last integral is clearly divergent.
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Nancy Phillips
Note that we have
$f\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}}\right\}=tu\left(t\right)$
and
$g\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}+{a}^{2}}\right\}=\frac{\mathrm{sin}\left(|a|t\right)}{|a|}u\left(t\right)$
Then, application of the convolution theorem yields
$\begin{array}{rl}{\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}\left({s}^{2}+{a}^{2}\right)}\right\}& =\left(f\ast g\right)\left(t\right)\\ \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(t-\tau \right)g\left(\tau \right)\phantom{\rule{thinmathspace}{0ex}}d\tau \\ \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left(t-\tau \right)u\left(t-\tau \right)\frac{\mathrm{sin}\left(|a|\tau \right)}{|a|}u\left(\tau \right)\phantom{\rule{thinmathspace}{0ex}}d\tau \\ \\ & ={\int }_{0}^{t}\left(t-\tau \right)\frac{\mathrm{sin}\left(|a|\tau \right)}{|a|}\phantom{\rule{thinmathspace}{0ex}}d\tau \end{array}$